Skip to main content

2022 AMC 12A Problem 15

Problem 15 of 25IntermediateAlgebraGeometry

The roots of the polynomial 10x339x2+29x610x^3-39x^2+29x-6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 22 units. What is the volume of the new box?

Answer choices

Show solution

Solution

Let the roots be r,s,t.r,s,t. By Vieta’s formulas, r+s+t=3910,r+s+t=\dfrac{39}{10}, rs+rt+st=2910,rs+rt+st=\dfrac{29}{10}, and rst=610=35.rst=\dfrac{6}{10}=\dfrac35. The new volume is (r+2)(s+2)(t+2)=rst+2(rs+rt+st)+4(r+s+t)+8=35+5810+15610+8=30. \begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered} Thus, the correct answer is D.

More practice

Concepts: Vieta’s Formulas · volume

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.